Problem: Find all complex numbers $z$ such that
\[z^2 = -77 - 36i.\]Enter all complex numbers, separated by commas.
Solution: Let $z = a + bi.$  Then
\[z^2 = (a + bi)^2 = a^2 + 2abi + b^2 i^2 = a^2 + 2ab - b^2.\]We want this to equal $-77 - 36i.$  Setting the real and imaginary parts equal, we get
\begin{align*}
a^2 - b^2 &= -77, \\
2ab &= -36,
\end{align*}so $ab = -18.$  Then $b = -\frac{18}{a}.$  Substituting, we get
\[a^2 - \frac{324}{a^2} = -77,\]so $a^4 + 77a^2 - 324 = 0.$  This factors as $(a^2 - 4)(a^2 + 81) = 0,$ so $a^2 = 4.$

If $a = 2,$ then $b = -\frac{18}{a} = -9.$  If $a = -2,$ then $b = -\frac{18}{a} = 9.$  Therefore, the solutions are $\boxed{2 - 9i, -2 + 9i}.$